BZOJ1270: [BeijingWc2008]雷涛的小猫

1270: [BeijingWc2008]雷涛的小猫

Time Limit: 50 Sec  Memory Limit: 162 MB
Submit: 1338  Solved: 707
[Submit][Status][Discuss]

Description

Input

Output

Sample Input

Sample Output

8

HINT

Source

dp水题

题解见代码注释

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(int &a, int &b)
{
    int tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const int INF = 0x3f3f3f3f;
const int MAXN = 5000 + 10;

//dp[i][j]表示当前高度为i，在第j颗树上的最大数量 
//ma[i]表示在高度为i时的最大数量
//dp[i][j] = max(dp[i - 1][j], ma[i - delta]) + val[i][j]
//ma[i] = max(ma[i], ma[j])

int dp[MAXN], ma[MAXN], val[MAXN][MAXN], n, h, delta;

int main()
{
    read(n), read(h), read(delta);
    for(register int i = 1;i <= n;++ i)
    {
        int tmp;read(tmp);
        for(register int j = 1;j <= tmp;++ j)
        {
            int tt;
            read(tt);
            ++ val[tt][i];
        }
    }
    int ans = 0;
    for(register int i = 1;i <= h;++ i)
        for(register int j = n;j >= 1;-- j)
        {
            if(i - delta >= 0) dp[j] = max(dp[j], ma[i - delta]);
            dp[j] += val[i][j];
            ma[i] = max(ma[i], dp[j]);
            ans = max(ans, dp[j]);
        }
    printf("%d", ans);
    return 0;
}